Question: $h(n) = 6n^{2}-n$ $g(t) = -t^{3}+5t^{2}+3t-h(t)$ $ h(g(2)) = {?} $
Explanation: First, let's solve for the value of the inner function, $g(2)$ . Then we'll know what to plug into the outer function. $g(2) = -2^{3}+5(2^{2})+(3)(2)-h(2)$ To solve for the value of $g$ , we need to solve for the value of $h(2)$ $h(2) = 6(2^{2})-2$ $h(2) = 22$ That means $g(2) = -2^{3}+5(2^{2})+(3)(2)-22$ $g(2) = -4$ Now we know that $g(2) = -4$ . Let's solve for $h(g(2))$ , which is $h(-4)$ $h(-4) = 6(-4)^{2}-(-4)$ $h(-4) = 100$